Need some log help (1 Viewer)

[Avenger6]

Hi, any help with the following questions would be greatly appreciated.

BTW, the 1 above the f in question 2a is supposed to indicate the derivative of, not to the power of 1.

 

[tommykins]

Are we trying to derive everything or?

 

[lyounamu]

What are we doing for q 1?

I got the answer for q 2

a)

f'(x) = (derivate of (√2-x ))/(√2-x)
= (-1/2 〖(2-x)〗^(-0.5) )/(√2-x)
= (-1)/(2 (2-x) )

= -1 / 4-2x
= 1/2x -4

b) f(x) = log10 x
= ln x / ln 10
f'(x) = 1/ln10 x

 

[foram]

Hi, any help with the following questions would be greatly appreciated.

BTW, the 1 above the f in question 2a is supposed to indicate the derivative of, not to the power of 1.

1a) Using Quotient Rule y'=(u'v-uv')/u^2

y'=(lnx . 2e^2x - (e^2x)/x)/ (lnx)^2

1b) y'=u'v+uv'
y'= e^x . lnx + (e^x)/x

2a) f(x) = ln (sqrt.(2-x))
EDIT: f(x) = 1/2 ln(2-x)
f'(x) = (1/2)(-1) / (2-x)
f'(x) = -1/ (4-2x)
f'(1) = -1/ (4-2)
= -1/2

2b) y=log x
change of base
y=lnx/ln10
y'= 1/xln10

 

[Mark576]

Uh, lyounamu, read Q2 again, it asks for f'(1). You only found the derivative:

f'(x) = 1/(2x-4) => f'(1) = 1/(2-4) = -1/2

EDIT:

2a) f(x) = ln (sqrt.(2-x))
f'(x) = 1/ (2-x)^1/2 . -1 (chain rule)
f'(x) = -1/ (2-x)^1/2

Nope. Your method would work for the derivative of log_ex but the general rule is that d/dx (log_ef(x)) = f'(x)/f(x).

 

[lyounamu]

Uh, lyounamu, read Q2 again, it asks for f'(1). You only found the derivative:

f'(x) = 1/(2x-4) => f'(1) = 1/(2-4) = -1/2

Thanks. Both I and Foram forgot about that. f'(1) = -1/2 (anyone knows how to do this, hopefully. Just substitute 1 into the derivative).

 

[lyounamu]

1a) Using Quotient Rule y'=(u'v-uv')/u^2

y'=(lnx . 2e^2x - (e^2x)/x)/ (lnx)^2

1b) y'=u'v+uv'
y'= e^x . lnx + (e^x)/x

2a) f(x) = ln (sqrt.(2-x))
f'(x) = 1/ (2-x)^1/2 . -1 (chain rule)
f'(x) = -1/ (2-x)^1/2
f'(1) = -1/ (2-1)^1/2
f'(1) = -1/ 1
f'(1) = -1

2b) y=log x
change of base
y=lnx/ln10
y'= 1/xln10

Isn't 2a) -1/2???

 

[foram]

Isn't 2a) -1/2???

Yea, I forgot to take the power of 1/2 down.

 

[lyounamu]

Uh, lyounamu, read Q2 again, it asks for f'(1). You only found the derivative:

f'(x) = 1/(2x-4) => f'(1) = 1/(2-4) = -1/2

EDIT:

Nope. Your method would work for the derivative of log_ex but the general rule is that d/dx (log_ef(x)) = f'(x)/f(x).

Seems to me 09ers like Foram and I cannot argue with professional like you.

 

[tommykins]

Don't stop trying to answer however, it'll only teach you more.

 

[Avenger6]

Thanks for the responses guys. Question 1a and b were asked to differentiate, sorry i didn't mention that. Im having a hard time interpreting some of the answers you guys gave so heres an image of where I got to with 1a and 1b, hopefully you can tell me where I have gone wrong.

Just for reference the textbook answer to 1a is (e^2x(2xlnx-1))/x(lnx)^2 and the answer to 1b is e^x(1/x+lnx)

Foram, I dont understand how you turned f(x) = ln (sqrt.(2-x)) into f(x) = 1/2 ln(2-x), where did the 1/2 come from??? I also don't quite understand the third to your answer for question 2b, i see how it becomes y=lnx/ln10 but dont understand how when derivived you, make it y'= 1/xln10???

Thank you for the help.

 

[tommykins]

1a) u' = 2e^2x not 2xe^(2x-1)
1b) u' = e^x, not xe^x-1

It isn't like normal differentiation with logs.

y = e^f(x)
y' = f'(x).e^f(x)


EDIT - can you tell me how you got that into a picture? Did you make it a PDF? If so, please PM me on how to do it

 

[cwag]

hey avenger 6...with q1 your on the right track...just u' is 2e^2x

so y' = (u'v-v'u)/v²

y'= {(2e^2x)(lnx)-(1/x)(e^2x)}/(lnx)²

by taking a common factor of e^2x out we get

y'= (e^2x(2lnx - 1))/x(lnx)²


for question 2
y=e^xlnx
y'= e^x(lnx) + (1/x)(e^x)
taking common factor out...
y'= e^x(1/x + lnx)

 

[foram]

Thanks for the responses guys. Question 1a and b were asked to differentiate, sorry i didn't mention that. Im having a hard time interpreting some of the answers you guys gave so heres an image of where I got to with 1a and 1b, hopefully you can tell me where I have gone wrong.

Just for reference the textbook answer to 1a is (e^2x(2xlnx-1))/x(lnx)^2 and the answer to 1b is e^x(1/x+lnx)

Foram, I dont understand how you turned f(x) = ln (sqrt.(2-x)) into f(x) = 1/2 ln(2-x), where did the 1/2 come from??? I also don't quite understand the third to your answer for question 2b, i see how it becomes y=lnx/ln10 but dont understand how when derivived you, make it y'= 1/xln10???

Thank you for the help.

For 1.
Your u' is wrong.

If y= e^x
y' = e^x

If y= e^f(x)
y' = f'(x).e^f(x) (chain rule)

Therefore if u=e^2x
u'=2e^2x

If u=e^x
u'=e^x

For 2.

sqrt.[f(x)] = [f(x)]^(1/2)

ln [f(x)]^a = a.ln[f(x)]

Therefore sqrt.(2-x) = (2-x) ^(1/2)

ln(2-x)^(1/2) = 1/2 .ln(2-x)

For 2 b)

For y=(ln[f(x)]/lna) ...[where a is a constant, lna must also be a constant]
therefore y' = [1/[f(x)] . f'(x)] /lna

 

[foram]

1a) u' = 2e^2x not 2xe^(2x-1)
1b) u' = e^x, not xe^x-1

It isn't like normal differentiation with logs.

y = e^f(x)
y' = f'(x).e^f(x)


EDIT - can you tell me how you got that into a picture? Did you make it a PDF? If so, please PM me on how to do it

Type it up in word 2007 then press alt + printscreen. Open paint, press ctrl +v. Cut out the part you want. copy and paste into new paint.

 

[Avenger6]

Thanks for the help guys, makes much more sense now . Tommykins, as foram said, i use word 07 and photoshop to produce and cut the equations. Thanks again.